# How many cardinals are there?

I'm trying to do the following exercise:

EXERCISE 9(X): Is there a natural end to this process of forming new infinite cardinals? We recommend this exercise instead of counting sheep when you have trouble falling asleep.

(This is from W. Just and M. Weese, Discovering Modern Set Theory, vol.1, p.34.)

By this process they mean $|\mathbb N| < |\mathcal P(\mathbb N)| < |\mathcal P (\mathcal P (\mathbb N))| < \dots$. My first response to this was "Obviously there is no end to it." but then the exercise is supposed to be challenging ("X-rated") so this must be wrong and there is an end to it. But when exactly? How many cardinals are there? What would be a "natural end"? Thank you for your help!

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205 :-) –  Brian M. Scott Oct 26 '12 at 20:01
@BrianM.Scott : ) –  Matt N. Oct 26 '12 at 20:02
@Brian: Any large cardinals in this collection? –  Asaf Karagila Oct 26 '12 at 23:19
@Asaf: Probably. Why, I’ll bet that a tailor could prove them to be measurable! –  Brian M. Scott Oct 26 '12 at 23:34
@Asaf: Although a quick glance shows that there is an almost-Erdős cardinal as well as (very surprisingly!) an almost-Reinhardt cardinal. –  Arthur Fischer Oct 27 '12 at 8:34

I don’t know what they have in mind, but the process as described goes only $\omega$ steps. To go further, you have take the union of what you already have and start over. That is, the process of repeatedly taking the power set will get you $\beth_0,\beth_1,\beth_2,\dots$ and hence $\beth_n$ for each $n\in\omega$, but it won’t get you beyond those. If you take the union of all those power sets, you get something whose cardinality is $\beth_\omega$, and you can start powering up again.

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I think this could be what they had in mind when they wrote it. Thank you! I didn't think of limit cardinals. –  Matt N. Oct 26 '12 at 20:07
@MattN.: You’re welcome! –  Brian M. Scott Oct 26 '12 at 20:08

I think what this exercise invites you to contemplate is what "a natural end" could even mean here.

It is clear that the $|\mathcal P^n(\mathbb N)|$ sequence gives you a countable infinity of different infinite cardinals, but is that an "end"? And if it is, would it be a "natural" one? "Countably infinite" is, as you know by now, a rather poor and sickly kind of infinity. Could we have continuum many different cardinals, for example?

If you have learned about initial ordinals, you might consider what this concept implies for the order type of the collection of all cardinals with their inherent order.

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Thank you! I don't remember having heard of initial ordinals but they might appear later in the book. –  Matt N. Oct 26 '12 at 20:09
The way I like to put it (and suggested to the author of a fairly well known real analysis text a few years ago, who put it in his book), given any cardinal number $b,$ there exists (a set containing) more than $b$ many cardinal numbers. For example, letting $b^{+}$ denote the successor cardinal of $b,$ then $\{\beth_{\alpha}:\; \alpha < b^{+} \}$ is a set containing exactly $b^{+}$ many cardinal numbers. –  Dave L. Renfro Oct 26 '12 at 20:30

Assume there is an end and $A$ is a set of maximal cardinality. Then you still have $|\mathcal P(A)|>|A|$, contradicton.

It is even worse: The "set" of cardinalities is not a set - because it could not possibly have a cardinality. (Though there is some longer story behind that).

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Thank you. I suspect what Brian says: they might want to hint at limit cardinals. –  Matt N. Oct 26 '12 at 20:10
@MattN. On seond reading of "this process" it seems you are right. –  Hagen von Eitzen Oct 26 '12 at 20:12

I think it is important to consider the context in which this question was posed.

We first have Cantor's Theorem (Theorem 5, p.32) which says that $| A | < | \mathcal{P} (A) |$ for any set $A$.

But we also have Theorem 7 (p.34), which I will quote in its entirety:

Let $\mathcal{M}$ be a set of cardinals such that $\forall \mathfrak{m} \in \mathcal{M} \exists \mathfrak{n} \in \mathcal{M} ( \mathfrak{m} < \mathfrak{n} )$, and let $\mathcal{A}$ be a family of sets such that $\forall \mathfrak{m} \in \mathcal{M} \exists A \in \mathcal{A} ( |A| = \mathfrak{m} )$. Then $| \bigcup \mathcal{A} |$ exceeds every cardinal in $\mathcal{M}$.

It is this last ingredient that allows this process to extend beyond the simple successor cases. (And this is what everyone else has been saying.)

Start with $X_0 , X_1 = \mathcal{P}(X_0) , X_2 = \mathcal{P} (X_1) , \ldots , X_{n+1} = \mathcal{P} ( X_n ) , \ldots$ and then after $\omega$ steps take $X_\omega = \bigcup_{n < \omega} X_n$. By Theorem $7$ we know that $| X_\omega | > | X_n |$ for all $n < \omega$. We continue with $X_{\omega + 1} , X_{\omega_2} , \ldots , X_{\omega \cdot 2} , X_{\omega \cdot 3} , \ldots , X_{\omega \cdot \omega}$ etc. We can even imagine this extending beyond the countable ordinals, arriving at $X_{\omega_1}$ which will satisfy $| X_{\omega_1} | > | X_\alpha |$ for all $\alpha < \omega_1$. And this can continue still more.

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This is a matter of philosophical approach to what it means "a natural end". After all, there is no natural end to the generation of natural numbers -- but certainly they end somewhere because we know that most ordinals are uncountable.

On the same note generation of cardinals does not end as long as you have ordinals to iterate through (either in the power set operation; or in the Hartogs number operation; or limits of these constructions). In this sense there is no natural end to the process of generating new cardinals.

However if one thinks of operations performed along the class of ordinals as operations which terminate when you "run out of ordinals", and if you can run out of natural numbers there is no reason you cannot run out of ordinals as well, in such case there is in fact a natural ending to the generation of new sets and therefore of new cardinals.

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